Proving Subsets: A Venn Diagram Approach

[leilei]

Proof subset?

Given three sets A, B, and C, set X = (A-B) U (B-C) U (C-A) and
Y = (A∩B∩C) complement C. Prove that X is subset of Y. Is Y necessarily a subset of X? If yes, prove it. If no, why?
---When I draw the two venn diagrams X and Y, they are the same, but I don't know how to prove it...

Can someone help me out here...
Thanks in advance!

 

[CompuChip]

The usual proof for such a statement is: Let x be an element from X and try to prove that it is also an element of Y. So if x is in X, you know that it is in A but not in B, and/or it is in B but not in C, and/or it is in C but not in A. You want to show that it must be in (A∩B∩C)^C. If it is in (A∩B∩C) then it would be in A and B and C, so it being in the complement means it is at least not in A or not in B or not in C. So you could suppose it is both in A and B and show that it cannot be in C.

So let me write this out in the right order:
Let x \in X. Suppose that x \in A, x \in B. Then definitely, x \not\in A - B, x \not\in C - A, because if it is in A it cannot be in any set from which we remove all elements of A (and similarly for B). But it must be in one or more of (A-B), (B-C) and (C-A), so it must be in (B - C). That is, x is in B (which we knew) but not in C. So if x is not in C, it cannot be in the intersection of C with whatever set you make up. In particular, it is not in A \cap B \cap C. Therefore, it must be in the complement of that set, which is called Y.

Now try to do the same reasoning for Y \subset X. You have already shown by your Venn diagram that if x lies in Y, it must lie in X. So try to prove it in the same way as I just did.

 

[leilei]

Thanks a lot !

 

[HallsofIvy]

There is, however, a technical problem with "Let x \in X"- the proof collapses is X is empty. Far better to start "IF x \in X". That way, if X is empty, the hypothesis is false and the theorem is trivially true.

 

[CompuChip]

You are right, obviously the empty set is a subset of any set S (vacuously, all its elements are also in S).